Answer
See the image below.
Work Step by Step
(a) To make our work easier we can first graph $y=e^x$ and then apply transformations to get the graph we are asked for. (You can see this graph, a parent graph, as light grey graph in the image).
Next we can get the graph of $y_1=\frac{e^x}{2}$ by shrinking the graph of the parent function vertically by the factor of $2$. (Shown as the blue graph in the image)
Now we can simply get $y_2=\frac{e^{-x}}{2}$ by reflecting $y_2$ about $y$-axis. (Red graph in the image)
Using the graphical addition (Explained in the section 2.7) we can sketch the graph of $cosh(x)=\frac{e^x+e^{-x}}{2}$ using the graphs of $y_1$ and $y_2$.
(b) To prove that $cosh(x)=cosh(-x)$ we can input both of these values and then compare them. (For a better understanding we can follow the same way we did in (a)).
$cosh(x)=\frac{e^x+e^{-x}}{2}$
$cosh(-x)=\frac{e^{-x}+e^{-(-x)}}{2}=\frac{e^{-x}+e^{x}}{2}$
It's clear that both of them have the same expression which means the same graph and so they are equal to each other.